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I wish to show $$\int_{0}^{\infty}\exp(te^x-x^2/2)\text{ d}x$$ for $t$ in some neighborhood of $0$ is divergent.

Is this a valid proof?

The Taylor series for $e^x$ is $$e^x = 1 + x + \dfrac{x^2}{2}+O(x^3) \geq\dfrac{x^2}{2}$$ since we are given $x \geq 0$. If we suppose $t > 0$, then of course, $$te^x \geq \dfrac{x^2}{2}$$ so that the exponent $$te^x - \dfrac{x^2}{2} \geq 0$$ and hence $$e^{te^x - x^2/2} \geq e^0 = 1$$ since $e^{\cdot}$ is non-decreasing, and hence $$\int_{0}^{\infty}\exp(te^x-x^2/2)\text{ d}x \geq \int_{0}^{\infty}1\text{ d}x = \infty\text{.}$$

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  • $\begingroup$ So far so good. But what about $t<0$? $\endgroup$
    – polfosol
    Sep 25 '16 at 5:44
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    $\begingroup$ From $e^x\ge\dfrac{x^2}2$ you get $te^x\ge\dfrac{x^2}2$, but you have only assumed that $t>0$ not that $t>1$... Please have a look at my answer below. $\endgroup$ Sep 25 '16 at 9:04
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  1. Assume $t>0$. Since $$ \lim_{x \to \infty}\frac{e^x}{\frac{x^2}2}=\infty $$ then there exists $x_0>0$ such that for all $x>x_0$, $$ \frac{e^x}{\frac{x^2}2}>\frac1t,\qquad te^x-\frac{x^2}2>0, $$ giving, for any $M>x_0$, $$ \begin{align} \int_{0}^{M}\exp(te^x-x^2/2)\:dx&=\int_{0}^{x_0}\exp(te^x-x^2/2)\:dx+\int_{x_0}^{M}\exp(te^x-x^2/2)\:dx \\\\&>\int_{0}^{x_0}\exp(te^x-x^2/2)\:dx+\int_{x_0}^{M}1\:dx \\\\&>M-x_0 \end{align} $$ yielding, as $M \to \infty$, the divergence of $\displaystyle \int_{0}^{\infty}\exp(te^x-x^2/2)\:dx$ in this case.
  2. Assume $t\le0$. Then $$ te^x-\frac{x^2}2\le -\frac{x^2}2, $$ giving $$ \begin{align} \int_{0}^{\infty}\exp(te^x-x^2/2)\:dx&\le\int_{0}^{\infty}\exp(-x^2/2)\:dx=\frac{\sqrt{\pi}}{\sqrt{2}},\end{align} $$ the given integral is convergent in this case.
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